SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Answer: no. Suppose that the segment $[0,1]$ can be represented as the union $A\cup B$ of two disjoint sets $A$ and $B$ such that $f(A)=B$. Let us immediately note that the function $f$ is strictly increasing on the interval $[0,1]$ and is a bijection from $[0,1]$ to $f([0,1])$. Let us prove that $f$ has no fixed points on the interval $[0,1]$. Indeed, let $f(s)=s$ for some $s\in [0,1]$. If $s\in A$, then $f(s)=s\notin B$, and if $s\in B$, then there is no number from $A$ that is sent to $s$, because $f$ is a bijection. Since both cases are impossible and $[0,1]=A\cup B$, we arrive at a contradiction, that is, there cannot be fixed points. Due to the continuity of the function, this means that $f(x)>x$ for all $x\in [0,1]$ or $f(x)<x$ for all $x\in [0,1]$ depending on the sign of $c$. Case $c\ge 0$. We have $x<f(x)$ for all $x\in [0,1]$. Since $f(0)=c>0$ and the function is increasing on the interval $[0,1]$, then $f(x)\ge c$ for any $x\in [0,1]$. Therefore $[0,c[\subseteq A$, whence $[c,f(c)][\subseteq B$, and $c<f(c)$. Let us now consider the interval $[f(c),f(f(c))[$. Since all points of this interval are images of points from $B$ and not from $A$ (after all, $f$ is a bijection), the intersection of $[f(c),f(f(c))[$ with $B$ is empty. We know that $f(c)\le 1$ and $f(c)<f(f(c))$. Note that $1\notin [f(c),f(f(c))[$, because otherwise we would get that $1\notin B$, whence $1\in A$ and $1<f(1)\in B$, which is impossible. Consequently, $[f(c),f(f(c))[$ $\subseteq A$ and, continuing the reasoning in this way, we obtain that the sequence $0,f(0),f^2(0),f^3(0),\dots$ increases and lies entirely in $[0,1]$, that is, it is bounded above. This means that it has a limit $\ell \in [0,1]$. From the continuity of $f$ we obtain that $$\ell =\underset{n\to \infty }{\lim }{f}^n(0)=f\left(\underset{n\to \infty }{\lim }{f}^{n-1}(0)\right)=f(\ell ),$$ that is, $\ell$ is a fixed point. Contradiction. Case $c<0$. We have $f(x)<x$ for all $x\in [0,1]$. In particular, $f(1)<1$. The function is increasing on the interval $[0,1]$, and so $f(x)\le f(1)$ for any $x\in [0,1]$. Therefore $]f(1),1]\subseteq A$, whence $]f(f(1)),f(1)]\subseteq B$, and $f(f(1))<f(1)$. Similar to the previous case, we show that $]f^3(1),f^2(1))[$ $\subseteq A$ and that the sequence $$1,f(1),f^2(1),f^3(1),\dots$$ converges to some number $\ell \in [0,1]$, which is a fixed point for $f$. Contradiction.